Understanding Joint Probability Distributions in Statistics

Joint probability distributions are crucial in analyzing the simultaneous behavior of random variables. They can be described using mass functions for discrete variables and density functions for continuous variables. This concept is fundamental in probability and statistics, aiding in calculating probabilities and making informed decisions in various fields.

• Probability
• Statistics
• Random Variables
• Distribution
• Probability Mass Function

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1. Joint Probability Distributions In general, if X and Y are two random variables, the probability distribution that defines their simultaneous behavior is called a joint probability distribution. 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

2. Note: If X and Y are 2 discrete random variables, this distribution can be described with a joint probability mass function. If X and Y are continuous, this distribution can be described with a joint probability density function. 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

3. Two Discrete Random Variables: If X and Y are discrete, with ranges ?? and ?? , respectively, the joint probability mass function is p(x, y) = P(X = x and Y = y), x ?? , y ??. 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

4. in the discrete case, 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

5. Two Continuous Random Variables: If X and Y are continuous, the joint probability density function is a function f(x,y) that produces probabilities: ? ?,? ? = ? ?,? ???? A 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

6. in the continuous case, b d ( ) = 4) , ( , ) f x y dydx P a x b c y d a c 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

7. Example: Suppose we have the following joint mass function Y -2 0 K 5 X 1 3 0.15 0.20 0.20 0.15 0.05 Find the value of k? 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

8. Answer: Using ?(?,?) = 1 ? ? We get 0.15 + 0.20 + ? + 0.05 + 0.20 + 0.15 = 1 0.75 + ? = 1 ? = 1 0.75 = 0.25 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

9. Example: Suppose we have the following joint density function 6 x y 0 2 , 2 4 x y = ( , ) f x y 8 0 . . OW 1) Prove that ?(?,?) is a joint probability function? 2) Calculate ? ? 2 3,? 5 2 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

10. Answer: 1) ( , f x y ) 0 6 x y 4 2 = ( , ) f x y dxdy 8 2 0 1 8 4 2 ( ) = 6 x y dx dy 2 0 2 1 8 2 x 4 = 6 x yx dy 2 2 0 1 8 (2) 2 2 4 = 6(2) (2) 0 y dy 2 1 8 110 8 1 40 16 8 4 ( ) = 10 2 y dy 2 ( ) ( ) 4 = = 10(4) (4) 10(2) (2) 2 2 2 y y 2 1 8 ( ) ( ) ( ) 8 = = = 20 4 1 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

11. 2 5 3 2 2 3 5 2 6 x y = 2) , P x y dydx 8 0 2 41 288 Prove that? = = 0.142 Another example see Ex 3.15 on page 96 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

12. The marginal distributions 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

13. Example: Suppose we have the following joint mass function Y -2 0 5 X 1 3 0.15 0.20 0.25 0.05 0.20 0.15 Find the marginal distributions of X and Y? 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

14. Answer: -2 0 5 Sum Y X 0.15 0.25 0.20 0.6 1 0.20 0.05 0.15 0.4 3 Sum 0.35 0.30 0.35 1 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

15. So x 1 3 Sum The marginal distribution of X 0.6 0.4 1 ( ) f x -2 0 5 Sum y ( ) f y 0.35 0.30 0.35 1 The marginal distribution of Y 15 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

16. Example: Suppose we have the following joint density function ( ) = + ( , ) f x y , 0 1, 0 2 c x y x y Find the value of c ? Find the marginal distributions of X and Y? 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

17. Answer: 2 1 ( ) = + = 1) ( , ) 1 1 f x y dxdy c x y dxdy 0 0 1 3 1 3 ( ) = = + ( , ) c f x y x y 1 3 2 ( ) = = + 2) ( ) ( , ) f x f x y x y dy 0 y 2 3 ( ) = + ( ) 1 f x x 1 3 1 ( ) = = + ( ) y ( , ) f f x y x y dx 0 x 1 3 1 2 = + ( ) y f y 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

18. conditional probability distribution 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

19. Example: 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

20. Solution: 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

21. Another example see Ex 3.20 on page 100 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

22. Statistical Independence 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

23. Example: Suppose we have the following joint distribution 3 , 0 , 0 3 e e x OW y x y = ( , ) f x y 0 , . . Prove that X and Y are independent? 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

24. = ( , ) f x y ( ) ( ) f y f x = = = 1) ( ) ( , ) f x y dy 3 3 3 3 f x e e dy e e dy x y x y 0 0 3 e y = = 0 1 = 3 .....................(1) e e e x x x 3 0 = = = 2) ( ) f y ( , ) f x y dx 3 3 3 3 e e dx e e dx x y y x 0 0 = = 0 1 = 3 3 3 ... ............(2) 3 3 3 e e e e y x y y 0 (1) (2) From and = ( , ) f x y ( ) ( ) f y f x 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

25. Notes: if X and Y are independent, then 1) ( , ) f x y = ( ) ( ) f y f x ( ( ) ) = 2) ( ) f x y f x = 3) ( ) f y f y x 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

26. Example: Suppose we have the following joint distribution ( ) ( , ) 8 f x y k x y = , 0 4 ,1 3 x y Find: 1)The value of k ( ( ) ( ) 2) ( ) , f x ( ) f y 3) , f y x f x y ) 4) ( 3) 5) 3 2 P x P x y 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

27. Solution: 3 4 = = 1) ( , ) 1 (8 ) 1 f x y dxdy k x y dxdy 1 0 3 4 = (8 ) 1 k x y dx dy 1 0 4 2 x 3 = 8 1 k x xy 2 1 0 2 4 2 3 = 8(4) 4 1 k y dy 1 3 ( ) + = 4 24 1 k y dy 1 3 2 4 y + = 24 1 k y 2 1 1 32 1 32 ( ) = = = 32 1 ( , ) (8 ) k k f x y x y 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

28. = 2) ( ) x ( , x y dy ) f f 1 32 3 ( ) = ( ) x 8 f x y dy 1 1 32 ( ) = ( ) x 12 2 , 0 4 f x x = ( ) y ( , x y dx ) f f 1 32 4 ( ) = ( ) y 8 f x y dx 0 1 32 ( ) = ( ) y 24 4 , 1 3 f y y 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

29. 1 32 ( ) 8 x y ( ) 8 x y ( , ) ( ) f x f x y ( ) = = = 3) f y x ( ) 1 32 12 2 x ( ) 12 2 x 1 32 1 32 ( ) 8 x y ( ) 8 x y ( , ) ( ) f y f x y ( ) = = = f x y ( ) 24 4 y ( ) 24 4 y 27 32 3 = = 4) ( 3) ( ) P x f x dx 0 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili

30. ( 3, 2) P x y ( ) = 5) 3 2 P x y ( 2) p y 1 32 30 64 2 3 2 3 ( ) = = = ( 3, 2) ( , ) f x y dxdy 8 P x y x y dxdy 1 0 1 0 1 32 18 32 2 2 ( ) = = 24 4 = ( 2) ( ) f y dy p y y dy 1 1 5 6 ( ) = 3 2 P x y 503 STAT - Probability and Statistics for Engineers and Scientists Dr. Mansour Shrahili